Write the equation of the line with x-intercept of 3 and y-intercept of -5, in point-slope form.

Increase £600 by 15%?

SIZIF [17.4K]

You will have to multiply 600 by 15%

600 x 0.15 = 90
600 + 90 = 690 Euro

8 0

3 years ago

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A sphere has a diameter of 12 meters. What is its exact surface area and its exact volume?

Karo-lina-s [1.5K]

Answer:

The exact value of its surface area = 144π m²

The exact value of its volume = 288π m³

Step-by-step explanation:

∵ The diameter of the sphere is 12m

∴ The radius of the sphere = 12 ÷ 2 = 6m

∵ The surface area of the sphere = 4πr²

∴ The surface area = 4 × π × 6² = 144π m²

∵ The volume of the sphere = 4/3 πr³

∴ The volume = $\frac{4}{3}\pi (6)^{3}$ = 288π m³

6 0

4 years ago

Sahana sells two products for ₹. 8000 each gaining 8% on one and losing 6% on the other. Find her gain or loss percent in the wh

Margaret [11]

Based on the information given the gain or loss percent on the whole transaction is 1%.

<h3>Gain or loss percent: </h3>

First step is to calculate the profit on the whole transaction

Profit=(8%×8,000)-(6%×8,000)

Profit=$640-$480

Profit=$160

Now let calculate the gain or loss percentage on the whole transaction

Gain or loss percentage=160/(8000+8000)×100

Gain or loss percentage=160/16000×100

Gain or loss percentage=1%

Inconclusion the gain or loss percent on the whole transaction is 1%.

Learn more about gain or loss here:brainly.com/question/25278228

5 0

2 years ago

Samantha wants to redecorate her room. She would like to replace the carpet and needs

CaHeK987 [17]

There are No pictures

6 0

3 years ago

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Find the point(s) on the surface z^2 = xy 1 which are closest to the point (7, 11, 0)

leonid [27]

Let $P=(x,y,z)$ be an arbitrary point on the surface. The distance between $P$ and the given point $(7,11,0)$ is given by the function

$d(x,y,z)=\sqrt{(x-7)^2+(y-11)^2+z^2}$

Note that $f(x)$ and $f(x)^2$ attain their extrema, if they have any, at the same values of $x$ . This allows us to consider the modified distance function,

$d^*(x,y,z)=(x-7)^2+(y-11)^2+z^2$

So now you're minimizing $d^*(x,y,z)$ subject to the constraint $z^2=xy$ . This is a perfect candidate for applying the method of Lagrange multipliers.

The Lagrangian in this case would be

$\mathcal L(x,y,z,\lambda)=d^*(x,y,z)+\lambda(z^2-xy)$

which has partial derivatives

$\begin{cases}\dfrac{\mathrm d\mathcal L}{\mathrm dx}=2(x-7)-\lambda y\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dy}=2(y-11)-\lambda x\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dz}=2z+2\lambda z\\\\\dfrac{\mathrm d\mathcal L}{\mathrm d\lambda}=z^2-xy\end{cases}$

Setting all four equation equal to 0, you find from the third equation that either $z=0$ or $\lambda=-1$ . In the first case, you arrive at a possible critical point of $(0,0,0)$ . In the second, plugging $\lambda=-1$ into the first two equations gives

$\begin{cases}2(x-7)+y=0\\2(y-11)+x=0\end{cases}\implies\begin{cases}2x+y=14\\x+2y=22\end{cases}\implies x=2,y=10$

and plugging these into the last equation gives

$z^2=20\implies z=\pm\sqrt{20}=\pm2\sqrt5$

So you have three potential points to check: $(0,0,0)$ , $(2,10,2\sqrt5)$ , and $(2,10,-2\sqrt5)$ . Evaluating either distance function (I use $d^*$ ), you find that

$d^*(0,0,0)=170$
$d^*(2,10,2\sqrt5)=46$
$d^*(2,10,-2\sqrt5)=46$

So the two points on the surface $z^2=xy$ closest to the point $(7,11,0)$ are $(2,10,\pm2\sqrt5)$ .

5 0

3 years ago